The sum of the digits of a two-digit number is 7. If we put the digits

1.	The sum of the digits of a two-digit number is 7. If we put the digits of the number in reverse order, the new number is 49 less than twice the original number. Find 40% of the number.
Answer» Step-by-step EXPLANATION: Here , let X and y be the two digits such that 10x+y is the number according to the question x+y= 7 x= 7-y (1) and again putting the digit in reverse i.e if 14(1×10+4) is the number, it would be 41 (4×10+1) 10y+x= 2(10x+y)-49 (2) putting the value of x in equation 2 10y+(7-y)=2[ 10(7-y)+y]-49 9y+7=2[70-10y+y]-49 9y+7= 140-18y-49 9y+18y= 91-7 27y= 84 9y= 28 please recheck your question

The sum of the digits of a two-digit number is 7. If we put the digits of the number in reverse order, the new number is 49 less than twice the original number. Find 40% of the number.

Answer»

Step-by-step EXPLANATION:

Here ,

let X and y be the two digits such that

10x+y is the number

according to the question

x+y= 7

x= 7-y (1)

and again putting the digit in reverse i.e if 14(1×10+4) is the number, it would be 41 (4×10+1)

10y+x= 2(10x+y)-49 (2)

putting the value of x in equation 2

10y+(7-y)=2[ 10(7-y)+y]-49

9y+7=2[70-10y+y]-49

9y+7= 140-18y-49

9y+18y= 91-7

27y= 84

9y= 28

please recheck your question

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The sum of the digits of a two-digit number is 7. If we put the digits of the number in reverse order, the new number is 49 less than twice the original number. Find 40% of the number.

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